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    How can \displaystyle\sum_{r=1}^n r = \frac{1}{2}n(n+1)
    be proven, without the usual writing the sequence in reverse and adding up.

    Also for \displaystyle\sum_{r=1}^n r^2 and \displaystyle\sum_{r=1}^n r^3.
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    (Original post by UKBrah)
    How can
    \displaystyle\sum_{r=1}^n r = \frac{1}

{2}n(n+1)
    be proven, without the usual writing the
    sequence in reverse and adding up.
    Also for \displaystyle\sum_{r=1}^n r^2[/

latex] and [latex]\displaystyle\sum_{r=1}^n

r^3.
    Are you asking about proof by induction?
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    Do you even lift brah?


    And see what you make of this Name:  sum5.png
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    (Original post by This Excellency)
    Are you asking about proof by induction?
    I suppose, but for PBI you're given the LHS and the RHS. When you're just given the summation how can you proof that it's equal to 1/2(n)(n+1) without the newfag method? Is it even possible?

    (Original post by marcus2001)
    Do you even lift brah?


    And see what you make of this Name:  sum5.png
Views: 51
Size:  1.4 KB
    revisingatthegym.jpeg

    no.
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    I wrote this without properly reading that this is sixthform but I'm posting it anyway just in case it might be of use to someone
    Let \sigma_{1}(n) = \sum\limits_{k=1}^{n}k, and \sigma_{2}(n) = \sum\limits_{k=1}^{n}k^2 and \sigma_{3}(n) = \sum\limits_{k=1}^{n}k^3, starting with the last and switching the order of summation we have

      \begin{aligned} \sigma_{3}(n) & = \sum\limits_{k=1}^{n}k^3 = \sum\limits_{k=1}^{n}\sum\limits  _{i=1}^{k}k^2 = \sum\limits_{i=1}^{n}\sum\limits  _{k=i}^{n}k^2 = \sum\limits_{i=1}^{n}\sum\limits  _{k=1}^{n}k^2-\sum\limits_{i=1}^{n}\sum\limits  _{k=1}^{i-1}k^2\end{aligned}

    Therefore  \sigma_{3}(n) = \sum\limits_{i=1}^{n}\sigma_{2}(  n)-\sum\limits_{i=1}^{n}\sigma_{2}(  i-1). Do the same thing for \sigma_{2}(n) then \sigma_{1}(n), and you will have all three sums.

    This really only relies on order switch (the third step). Pretty much nothing else. In general it lets you write \sigma_{r}(n) = \sum_{k=1}^{n}k^r in terms of \sigma_{r-1}(n), ~ \sigma_{r-2}(n), ~ \sigma_{r-3}(n), ~ \cdots, ~ \sigma_{0}(n) where of course \sigma_{0}(n) = n. (Shifting the index does the same thing for odd values of r).
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    (Original post by L'art pour l'art)
    I wrote this without properly reading that this is sixthform but I'm posting it anyway just in case it might be of use to someone
    Let \sigma_{1}(n) = \sum\limits_{k=1}^{n}k, and \sigma_{2}(n) = \sum\limits_{k=1}^{n}k^2 and \sigma_{3}(n) = \sum\limits_{k=1}^{n}k^3, starting with the last and switching the order of summation we have

      \begin{aligned} \sigma_{3}(n) & = \sum\limits_{k=1}^{n}k^3 = \sum\limits_{k=1}^{n}\sum\limits  _{i=1}^{k}k^2 = \sum\limits_{i=1}^{n}\sum\limits  _{k=i}^{n}k^2 = \sum\limits_{i=1}^{n}\sum\limits  _{k=1}^{n}k^2-\sum\limits_{i=1}^{n}\sum\limits  _{k=1}^{i-1}k^2\end{aligned}

    Therefore  \sigma_{3}(n) = \sum\limits_{i=1}^{n}\sigma_{2}(  n)-\sum\limits_{i=1}^{n}\sigma_{2}(  i-1). Do the same thing for \sigma_{2}(n) then \sigma_{1}(n), and and you will have all three sums.

    This really only relies on order switch (the third step). Pretty much nothing else. In general it lets you write \sigma_{r}(n) = \sum_{k=1}^{n}k^r in terms of \sigma_{r-1}(n), ~ \sigma_{r-2}(n), ~ \sigma_{r-3}, ~ \cdots, ~ \sigma_{0}(n) where of course \sigma_{0}(n) = n. (Shifting the index does the same thing for odd values of r).
    Sweet.
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    For just the summation of r, consider a graph with n number of nodes and you wish to calculate the number of arcs that connect all the nodes together. Still with me? Now use the sum of an arithmetic series which is 1/2n(2a+d(n-1)), you know that the first term in essence is n, so a=n and the common difference will be -1 because you are losing a node every time, and the last term stays in terms of n, tidy that up to give the summation formula for r. That is one way of looking at it, no idea how to approach the rest.
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    (Original post by 3nTr0pY)
    Sweet.
    Cheers! :]

    (Original post by Keynesian)
    ...
    Cool... but a question?
    Since \sum_{k=0}^{n}(a+bk)  = (n+1)a+b\sum_{k=1}^{n}k any derivation of the formula for an arithmetic series is essentially a calculation of \sum_{k=1}^{n}k?

    I've a feeling though your argument can (combinatorially) be used to calculate the sum without using the formula, but I don't know enough. :[

    If it's possible it's cooler and more beautiful that way (otherwise why introduce graphs rather than just plugging the values in the formula ).
 
 
 
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